多键快速排序崩溃

Question

我正在写多键快排，我编译好，但是当我运行，停止工作异常，我认为它循环永远或类似这样的事情，我怎么能解决这个问题呢？是链接，它 http://ideone.com/bBtaX

写入运行时错误，在这里也代码

#include<stdlib.h> 
#include<stdio.h> 
#include<string.h> 
#include<iostream> 

using namespace std; 

int min(int a,int b){ 
    return a<=b?a:b; 
} 

#define swap(a,b){ char *t=x[a];x[a]=x[b];x[b]=t;} 
#define i2c(i) x[i][depth] 

void vecswap(int i, int j, int n, char *x[]) 
{ 
    while (n-- > 0) { 
     swap(i, j); 
     i++; 
     j++; 
    } 
} 

void ssort1(char *x[],int n,int depth); 

void ssort(char *x[],int n) 
{ 
    ssort1(x,n,0); 
} 

void ssort1(char *x[],int n,int depth){ 
    int a,b,c,d,r,v; 
    if(n<=1) 
     return ; 
    a=rand()%n; 
    swap(0,a); 
    v=i2c(0); 
    a=b=1; 
    c=d=n-1; 
    for (;;) 
    { 
     while(b<=c && (r=i2c(b)-v)<=0){ 
      if (r==0) { 
       swap(a,b);a++; 
      } 
      b++; 
     } 

     while(b<=c && (r=i2c(c)-v)>=0){ 
      if (r==0) { 
       swap(c,d); d--; 
      } 
      c--; 
     } 

     if (b>c) 
      break; 
     swap(b,c); 
     b++; 
     c--; 
    } 
    r=min(a,b-a); 
    vecswap(0,b-r,r,x); 
    r = min(d-c, n-d-1); 
    vecswap(b, n-r, r, x); 
    r=b-a; 
    ssort1(x,r,depth); 
    if (i2c(r)!=0) 
     ssort1(x+r,a+n-d-1,depth+1); 
    r=d-c; 
    ssort1(x+n-r,r,depth); 
} 

int main(){ 
    char *s[]={"dato","giorgi","beso","computer","deda","kata","spilo"}; 
    int n=sizeof(s)/sizeof(char); 

    ssort(s,n); 
    for (int i=0;i<n;i++) 
     cout<<s[i]<<" "<<endl; 
    return 0; 
} 

Answer 1

此：

int n=sizeof(s)/sizeof(char); /* Would return 28, resulting in out of bounds 
           on array 's' in subsequent for loop. */ 

应该是：

int n=sizeof(s)/sizeof(char*); 
         ^

确定数组中的元素的数量的更安全的方法是：

int n=sizeof(s)/sizeof(s[0]);